Question

A factory manufactures two types of screws A and B, each type requiring the use of two machines, an automatic and a hand-operated. It takes 4 minutes on the automatic and 6 minutes on the hand-operated machines to manufacture a packet of screws ‘A’ while it takes 6 minutes on the automatic and 3 minutes on the hand-operated machine to manufacture a packet of screws ‘B’. Each machine is available for at most 4 hours on any day. The manufacturer can sell a packet of screws ‘A’ at a profit of 70 paise and screws ‘B’ at a profit of < 1. Assuming that he can sell all the screws he manufactures, how many packets of each type should the factory owner produce in a day in order to maximize his profit ? Formulate the above LPP and solve it graphically and find the maximum profit.

Answer 1

Answer:

Step-by-step explanation:

Let number of packets of type A = x

and number of packets of type B = y

∴ L.P.P. is: Maximize, Z = 0.7x + y 1

subject to constraints:

4x + 6y ≤ 240 or 2x + 3y ≤ 120

2

6x + 3y ≤ 240 or 2x + y ≤ 80

x ≥ 0, y ≥ 0

Correct graph 2

Z(0, 0) = 0, Z(0, 40) = 40

Z(40, 0) = 28, Z(30, 20) = 41 (Max.)

∴ Max. profit is ` 41 at x = 30, y = 20.